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13.Statistics
hard
A data consists of $n$ observations
${x_1},{x_2},......,{x_n}.$ If $\sum\limits_{i - 1}^n {{{({x_i} + 1)}^2}} = 9n$ and $\sum\limits_{i - 1}^n {{{({x_i} - 1)}^2}} = 5n,$ then the standard deviation of this data is
A
$5$
B
$\sqrt 5$
C
$\sqrt 7$
D
$2$
(JEE MAIN-2019)
Solution
${\sum {\left( {{x_i} + 1} \right)} ^2} = 9n\,\,\,\,\,\,….\left( 1 \right)$
${\sum {\left( {{x_i} – 1} \right)} ^2} = 5n\,\,\,\,\,\,….\left( 2 \right)$
$\left( 1 \right) + \left( 2 \right) \Rightarrow \sum {\left( {x_i^2 + 1} \right)} = 7n$
$ \Rightarrow \frac{{\sum {x_i^2} }}{n} = 6$
$\left( 1 \right).\left( 2 \right) \Rightarrow 4\sum {{x_i}} = 4n$
$ \Rightarrow \sum {{x_i} = n} $
$ \Rightarrow \frac{{\sum {{x_i}} }}{n} = 1$
$ \Rightarrow $ variance $=6-1=5$
$ \Rightarrow $standard diviation $ = \sqrt 5 $
Standard 11
Mathematics
